3
$\begingroup$

In Spivak's "A Comprehensive Introduction to Differential Geometry" Spivak defines a vector bundle as a tuple: $(E, \pi, B, \bigoplus, \bigodot),$ where $E$ is the total space, $B$ is the base space, $\pi$ is a continuous map from $E$ onto $B$ (thought of as like a projection) and there are vector additions on each fiber, and scalar multiplication on the total space. Lastly, he gives a local trivialization condition.

He says that two vector bundles over the same space are equivalent if there is a homeomorphism of the total spaces which takes fibers onto fibers. This is what he says is the definition of equivalence.

In an exercise after this, he says that we do not need a homeomorphism - that if the map between total spaces is continuous, then the spaces are in fact homeomorphic.

I don't even begin to see what's going on here. For manifolds in general, this is far from true - there is no reason continuity should say anything about the inverse map being continuous, and certainly not bijectivity. The only thing I can think of is that Spivak's question is worded a little poorly, and I should assume the map is continuous, and the fibers are taken isomorphically to fibers. Even with this new information though, I still feel mostly lost.

For example, why can't we send the whole total space to just a single point and fiber?

Thanks for any pointers or hints.

$\endgroup$
  • $\begingroup$ Take a base space B and two trivial vector bundles on B of ranks one and two. The inclusion map of one dimensional vector space into a two dimensional vector space (as a subspace) will induce a continuous map on vector bundles. Obviously it is not a homeomorphism. On the other hand if you look at a continuous map between two vector bundles of same rank then local triviality makes any continuous map a local homeomorphism. $\endgroup$ – DBS Feb 18 '16 at 10:42
  • $\begingroup$ @DBS Okay. I've thought about your comment for a little bit. I can see how it works kind of... if I already knew about bijectivity. We can piece together the local homeomorphisms into a global one. My problem is with the bijectivity though. We are only assuming that the map is continuous to itself. Of course it is surjective onto its image, but why is that image not a proper subset of $B$? (Specific example to follow).... (1/2) $\endgroup$ – Alfred Yerger Feb 18 '16 at 15:27
  • $\begingroup$ Consider for example a trivial bundle over $\mathbb R$. Over every fiber, we just have another copy of $\mathbb R$. In other words, we're looking at the plane. Take the map $f(x) = |x|$ on the base space. This map is continuous. Then map the fibers with just identity to whatever vector space they land on. In other words we'll extend $f$ onto $E$ by writing $f(x, y) = f(|x|, y)$. This map is continuous as a function of two variables, so it is continuous on $E$, but it's definitely not a homeomorphism. (2/2) $\endgroup$ – Alfred Yerger Feb 18 '16 at 15:29
4
$\begingroup$

Though the statement you mention above is somewhat loosely phrased let us try to understand what it means.

Notation: Our definition of a vector bundle is a tuple $(E,B, \pi)$ (I am ignoring the other components in your definition as they aren't relevant to us.) By local triviality I can write down points in the total space $E$ with co-ordinates $(b,t)$ where $b \in B$ and $t \in F_{b}$ where $F_{b}$ is the fiber over $b$.

We consider the case of two bundles $E, E^{\prime}$ and a map of the total spaces $f: E \rightarrow E^{\prime}$ realizing this equivalence. The underlying base space is $B$. In this case the equivalence is described by :

Definition 1 "He says that two vector bundles over the same space are equivalent if there is a homeomorphism of the total spaces which takes fibers onto fibers. This is what he says is the definition of equivalence. "

Consider the following alternate definition for equivalence of vector bundles.

Definition 2 An equivalence of vector bundles is an arrow $\varphi : (E, B, \pi) \rightarrow (E^{\prime}, B, \pi^{\prime})$ which induces (1) a continuous map $\varphi: E \rightarrow E^{\prime}$ that is an isomorphism (vector space) on the fibers and (2) the induced maps from $E$ to $B$, $\pi^{\prime} \varphi$ and $\pi$ are equal.

If we were given a $\varphi$ as above, let $\varphi_{b}$ denote the induced linear isomorphism between the fibers $F_{b}$ and $F^{\prime}_{b}$ over the point $b \in B$. Consider the function $\varphi^{-1}: E^{\prime} \rightarrow E$ defined at a point $(b,t) \in E^{\prime}$ as follows: $$\varphi^{-1}(b,t) = (b, \varphi_{b}^{-1}(t)).$$

we claim that $\varphi^{-1}$ is a continuous map of topological spaces- this follows again by local triviality. Hence $\varphi^{-1}$ defines an inverse of vector bundle map $\varphi$ according to definition 2.

Let us see the equality of these definitions.

$(2 \implies 1)$

By local triviality any point $e\in E$ has a co-ordinate $(b,t)$ where $b\in B$ and $t$ is in the fiber. If we were given a map $\varphi$ according to definition (2) we define $f(b,t) = (b, \varphi_{b}(t))$. This makes $f$ into a continuous and bijective map. We consider now the map $g: E^{\prime} \rightarrow E$ by $g(b,t) = (b, \varphi^{-1}_{b}(t))$. By the same argument $g$ is a continuous bijective map and hence it makes $f$ into a homeomorphism.

$(1 \implies 2)$ In this case we define $\varphi(b,t) = f(b,t)$. In this case everything is simpler except perhaps the condition $\pi^{\prime} \varphi = \pi$. To this end, given a vector bundle $(E,B, \pi)$ we have a continuous map of topological spaces $z: B \rightarrow E$ given by $z(b) = (b,0)$. If you aren't sure check this is continuous -local triviality. So the map composite map of topological spaces $$ B \stackrel{z}{\rightarrow} E \stackrel{f}{\rightarrow} E^{\prime} \stackrel{\pi^{\prime}}{\rightarrow} B$$ is identity and so is $\pi\circ z: B \rightarrow B$. This gives the required equality.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Your second condition dispenses with my counter-example. I don't believe this part is in Spivak... And it is the key. I understand now. $\endgroup$ – Alfred Yerger Feb 18 '16 at 18:39
  • $\begingroup$ It is my personal opinion that the first three chapters of Milnor's characteristic classes is a much better intro to vector bundle theory. maths.ed.ac.uk/~aar/papers/milnstas.pdf If you are categorically oriented then you should think of vector bundle as a category of vector spaces over a base (the base B). Morphisms in this category are morphisms of objects (vector bundles) that agree on the base (this is the second condition...). I recommend the categorical way of thinking (which is from Milnor) in this case. Because a natural question is how to compare vector bundles.... $\endgroup$ – DBS Feb 18 '16 at 19:29
  • $\begingroup$ ....over different bases, how to modify vector bundles, pretty soon there will be more general fiber bundles. The whole point of these concepts is to relate geometry on complicated objects (the total space) into complicated (but conceptually simpler) geometry (linear algebra in this case) over simple objects (the base for example). So it is useful to understand morphisms from this perspective... $\endgroup$ – DBS Feb 18 '16 at 19:32
  • $\begingroup$ I'll look into getting a copy from my local library. I am using Spivak as it is the book for the course, and while mostly classical, it is a friendly book that is easy to read. I'm using Lee's book in parallel, which I find to be more serious. Thank you for your help. $\endgroup$ – Alfred Yerger Feb 18 '16 at 22:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.